Constant Velocity on Inclined Plane Trick

 

❓ Question

A block of mass $m$ is on a $30^\circ$ inclined plane with coefficient of kinetic friction

$$\mu=\frac{\sqrt3}{2}.$$

Find the external force $F$ required so that the block moves with constant velocity along the incline.

(Given in the figure: $m=50\,\text{kg},\ g=10\,\text{m/s}^2$)

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✍️ Short Explanation

Since the block moves with constant velocity,

$$a=0$$

Hence, the net force along the incline must be zero.

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🔷 Step 1 — Forces Along the Incline

Along the incline:

• Component of weight (down the plane):

$$mg\sin30^\circ$$

• Kinetic friction (opposes motion):

$$f_k=\mu N$$

• Applied force $F$.

Normal reaction:

$$N=mg\cos30^\circ.$$

Therefore,

$$f_k=\mu mg\cos30^\circ$$
$$=\frac{\sqrt3}{2}\times mg\times\frac{\sqrt3}{2} =\frac34\,mg.$$

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🔷 Step 2 — Apply Equilibrium Along the Incline

For constant velocity,

$$F+f_k=mg\sin30^\circ.$$

Hence,

$$F = mg\sin30^\circ-f_k$$
$$= \frac12mg-\frac34mg = -\frac14mg.$$

The negative sign means the assumed direction of $F$ is incorrect.

So the required force has magnitude

$$\boxed{\frac{mg}{4}}$$

and acts opposite to the assumed direction.

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🔷 Step 3 — Numerical Value

Given,

$$m=50\ \text{kg},\qquad g=10\ \text{m/s}^2,$$
$$F=\frac{50\times10}{4}=125\ \text{N}.$$

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✅ Final Answer

$$\boxed{125\ \text{N}}$$

Direction: Down the incline (opposite to the initially assumed direction).

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🔷 JEE Shortcut

Compute directly:

$$F = mg\left(\sin30^\circ-\mu\cos30^\circ\right)$$
$$= mg\left(\frac12-\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}\right) = mg\left(\frac12-\frac34\right) = -\frac{mg}{4}.$$

So the force must be applied in the opposite direction, with magnitude

$$\boxed{\frac{mg}{4}=125\ \text{N}}.$$

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